问题
Was going through a Python Tutorial and came across an example to check whether a number is prime or not. I changed a few things so that the result would display a list of all possible factors of the number, if the number is not a prime, However, the code didn't work.
Code:
def isprime(number):
print "Reticulating Splines..."
fnum=[1,]
notprime=0
for p in range(2, number):
if (number % p) == 0:
notprime=1
fnum.append(p)
continue
if notprime == 1:
return number, "is not a prime because of these factors", fnum
else:
return True
num =int(raw_input("Enter number: "))
print isprime(num)
Output:
Enter number: 12
Reticulating Splines...
(12, 'is not a prime because of these factors', [1, 2, 3, 4])
>>>
Enter number: 25
Reticulating Splines...
(25, 'is a prime number')
Expected Output:
Enter number: 12
Reticulating Splines...
(12, 'is not a prime because of these factors', [1, 2, 3, 4, 6])
Enter number: 25
Reticulating Splines...
(25, 'is not a prime because of these factors', [1,5])
Poor control structure is my guess, but can someone fix my code?
I understand how range() works: In this case range() is given a start and a stop value, step defaults to 1. I understand that continue, continues a loop, but can I use it with if? I think that is wrong.
UPDATE
Solved, problem with indentation the continue should have been for the for loop, ditto for the if...notprime.
def isprime(number):
print "Reticulating Splines..."
fnum=[1,]
notprime=0
for p in range(2, number):
if (number % p) == 0:
notprime=1
fnum.append(p)
if notprime == 1:
return number, "is not a prime because of these factors", fnum
else:
return number, "is a prime number"
num =int(raw_input("Enter number: "))
print isprime(num)
Update2: (Thx to @neil)
And the continue is plain stupid
Updated Code and speed comparisons between n/2 and sqrt(n)
Thanks to @neil and @emmanuel
n/2 code: v2
import time
def isprime(number):
start=time.clock()
print "Reticulating Splines..."
fnum=[1,]
notprime=0
for p in range(2, (number/2)+1):
if (number % p) == 0:
notprime=1
fnum.append(p)
end=time.clock()
if notprime == 1:
return number, "is not a prime because of these factors", fnum, "Time taken", end-start
else:
return number, "is a prime number. Time Taken", end-start
print "Prime or factor calculator v2 using n/2"
print #
num =int(raw_input("Enter number: "))
print isprime(num)
sqrt(n) code: v3
import math, time
def isprime(number):
start=time.clock()
print "Reticulating Splines..."
fnum = [1,]
last = int(math.ceil(math.sqrt(number)))
for p in range(2, last + 1):
if (number % p) == 0:
fnum.append(p)
fnum.append(number / p)
# Remove duplicates, sort list
fnum = list(set(fnum))
fnum.sort()
end=time.clock()
if len(fnum) > 1:
return number, "is not a prime because of these factors", fnum ,"Time taken", end-start
else:
return True, "Time taken", end-start
print "Prime or factor calculator v3 using sqrt(n)"
print #
num =int(raw_input("Enter number: "))
print isprime(num)
Output(Showing only time)
Time for sqrt(n) code: v3
Prime or factor calculator v3 using sqrt(n)
Enter number: 999999
Time taken', 0.0022617399697466567
Time for n/2 code: v2
Prime or factor calculator v2 using n/2
Enter number: 999999
Time taken: 0.11294955085074321
Time for original code(n): v1
Prime or factor calculator v1
Enter number: 999999
Time taken: 0.22059172324972565
v1 and v2 could not handle numbers 999999999, 999999999999 and 999999999999999, both gave a MemoryError
However v3 handled the three numbers:
999999999 : 0.010536255306192288
999999999999 : 0.75631930873896636
999999999999999 : 24.04511104064909
The shell hangs for 9999999999999999 and gives a MemoryError for 999999999999999999
Thanks to @Lennart, I am thinking of rewriting the code in a more OOP friendly way, by using classes. But I don't seem to be doing it right.
回答1:
The problem is your indentation - if notprime==1: shouldn't be within the for loop. It should only have one level of indentation.
Also, the continue is unnecessary.
EDIT:
An improvement you can make (I was just working on primes last night for a Project Euler problem) is to only loop to n/2 - there can't be a factor greater than half the number.
回答2:
Your final if notprime ... and the three lines following it are indented too far and as such is executed inside the loop, instead of outside.
回答3:
You return after testing only one number. Move the if tests and the returns to outside of the for loop.
Also, returning True if it is a prime, and a string if it is not is unpractical.
If you then call
if isprime(7):
That will always evaluate as True. I've improved things a bit:
def factors(number):
fnum=[]
for p in range(2, number):
if (number % p) == 0:
fnum.append(p)
return fnum
for x in range(100):
f = factors(x)
if f:
print x, "is not a prime and has factors", f
else:
print x, "is a prime"
回答4:
@neil:
"An improvement you can make ... is to only loop to n/2 - there can't be a factor greater than half the number."
By the way, the highest value you need to test is int(math.ceil(math.sqrt(n))), no need to go to n/2 if each time you get a value, you get the associated one (i.e. if a x b = n, either a or b is lower than the square root of n and the other one is greater):
def isprime(number):
print "Reticulating Splines..."
fnum = [1,]
last = int(math.ceil(math.sqrt(number)))
for p in range(2, last + 1):
if (number % p) == 0:
fnum.append(p)
fnum.append(number / p)
# Remove duplicates, sort list
fnum = list(set(fnum))
fnum.sort()
if len(fnum) > 1:
return number, "is not a prime because of these factors", fnum
else:
return True
Greater performance, even if at the end the list is not sorted (but this can be done inside the loop by adding the 2 numbers p and number / p at the good indexes).
来源：https://stackoverflow.com/questions/4781841/calculate-all-possible-factors-of-a-prime